A square loop $ABCD$, carrying a current $i,$ is placed near and coplanar with a long straight conductor $XY$ carrying a current $I,$ the net force on the loop will be
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Force on arm $AB$ due to current in conductor $- XY$ is
$F_{1}=\frac{\mu_{0}}{4 \pi} \frac{2 l i L}{(L / 2)}=\frac{\mu_{0} I i}{\pi}$
acting towards $-17$ in the plane of loop.
Force on arm $CD$ due to current in conductor $XY$ is
$F_{2}=\frac{\mu_{0}}{4 \pi} \frac{2 I i L}{3(L / 2)}=\frac{\mu_{0} I i}{3 \pi}$
acting away from $-ill$ in the plane of loop.
$\therefore$ Net force on the loop $=F_{1}-F_{2}$
$=\frac{\mu_{0} I i}{\pi}\left[1-\frac{1}{3}\right]=\frac{2}{3} \frac{\mu_{0} I i}{\pi}$
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